Question

当使用更大的数组维度时，我已经将程序的终止隔离为始终在2D顶点顶点数据的广度搜索期间：

我的Vertex类被声明为

    class Vertex{                                                   
public:
    int i, j; 
    std::set<Vertex*> adj; //references to adjacent vertices (max of 4)

    //used in solving maze
    bool visited; 
    std::list<Vertex> path; //stores path from start vertex to this vertex

    Vertex();
    ~Vertex();
    //end constructors

    void setPos(int row, int col);

    /** for iterators */
    typedef std::set<Vertex*>::iterator iterator;
    iterator begin();
    iterator end();

};//END class Vertex

在这个函数中，我有这个小节来执行BFS：

void solveMaze(string folder, Vertex arr[][MAZE_WIDTH]){
    //<randomly generate begin and end positions in maze>
    //<ensure all vertices are marked unvisited>

/* BFS */
list< Vertex > shortestPath; //shortest path from start to end
queue<Vertex*> q; //store visited vertices with one or more unvisited adjacent vertices while searching graph

arr[ begini ][ beginj ].visited = true; //mark start vertex as visited, as the search will begin here
arr[ begini ][ beginj ].path.push_back( arr[ begini ][ beginj ] ); //all paths begin at start vertex
q.push( &arr[begini][beginj] );

//explore all vertices in maze
while( !q.empty() ){ 
    Vertex* cur = q.front(); //next vertex to backtrack
    q.pop();

    //for all vertices adjacent of cur
    Vertex::iterator it; //thanks to typedef
    for(it = (cur->begin()); it != (cur->end()); it++){
        //if adjacent vertex has not been visited
        if( !(*it)->visited ){ 
             (*it)->visited = true; //mark vertex visited
             (*it)->path = cur->path; //save directory from start vertex..
             cout << "1\n";
             ((*it)->path).push_back( **it ); //..to this vertex     i think this line is main problem
             cout << "2\n";
             q.push( &arr[ ((*it)->i) ][ ((*it)->j) ]); //current path continues, so store vertex to backtrack later
        }
    }

    //path to end found, so store it if it's shortest found
    if( (cur->i) == endi && (cur->j) == endj ){
        if( shortestPath.empty() )
            shortestPath = ( cur->path );
        else if( (cur->path).size() < shortestPath.size() )
            shortestPath = cur->path;
    }

}//end while( !q.empty() )

cout << "    -finished breadth first searching\n";
...

所以在BFS期间，1 2 1 2 ...重复打印，但是在最后1次打印之前越来越慢，然后它会终止并出现如下错误：

...
1
2
1
2
1

This application has requested the Runtime to terminate it in an unusual way.
Please contact the application's support team for more information.
make: *** [makeExec] Error 3

所以我怀疑这是我用指针/迭代器使用等做错的事情，比如我在上面的while循环中的行“（（* it） - ＆gt; path）.push_back（** it）;”因为1最后打印然后它终止，但它只在使用更大的数组维度时发生。

有人可以帮我弄清楚发生了什么吗？

Answer 1

更改

std::list<Vertex> path; //stores path from start vertex to this vertex

到

std::list<Vertex*> path; //stores path from start vertex to this vertex

工作得很漂亮。应该知道我浪费的资源，谢谢@uesp！

广度优先搜索中的c ++迭代器和指针用法

1 个答案: